Sub-divided and percentage bar diagrams
Farmers in two villages grew rice and pulses. If we look only at kilograms, the bigger village always seems bigger. But when we ask, "what percent of the total crop is rice?", the two villages can be compared fairly.
Sub-divided bar diagram: we add two or more things, draw one bar of the total height, and divide it into parts in proportion. Percentage bar diagram: every bar is taken as 100% tall and the parts are shown in percent.
In the figure: The same data in three ways. In the joint bar the parts are side by side, in the sub-divided bar one above the other, and in the percentage bar all bars are equally tall.
Rule 1: Sub-divided bar: add first, then divide in proportion — Add both numbers. Take the height of the total as per the scale. The lower part is the first number and the upper part is the second.
Sub-divided and percentage bar diagrams
In the app you can move this figure with your finger.
Open in the app- Easy: Rice 30 quintal and pulses 10 quintal. Total 40. Scale 1 cm = 10 quintal: the bar is 4 cm, the lower part 3 cm and the upper part 1 cm.
- Medium: Cow milk 45 litres and buffalo milk 15 litres. Total 60. Scale 1 cm = 10 litres: the bar is 6 cm, the lower part 4.5 cm and the upper part 1.5 cm.
- Exam type: Production of mango and cashew (quintal): year 1: 24 + 16 = 40; year 2: 30 + 30 = 60; year 3: 36 + 24 = 60. Bars of 4 cm, 6 cm, 6 cm. Years 2 and 3 have the same total, but mango has a bigger share in year 3.
Rule 2: Percentage bar: part ÷ total × 100 — Even if the yearly totals differ, take every bar as 100% tall. Find the percent of the first part; the upper part = 100 − that percent.
- Easy: 12 of 40 students are girls: 12 ÷ 40 × 100 = 30%. So boys are 70%.
- Medium: Rice 30 and pulses 10: total 40. Rice 30 ÷ 40 × 100 = 75% and pulses 25%.
- Exam type: Rice 30, 34, 40 and pulses 10, 16, 10 (three years). Totals 40, 50, 50. Rice 75%, 68%, 80%. Pulses 25%, 32%, 20%. The rice share is highest in the third year.
Rule 3: Choose the diagram from the data — Simple for one kind of data, joint for comparing two things, sub-divided for a total and its parts, and percentage when we want the percent of each part.
In the figure: The diagram that suits each kind of data is on the right.
- Easy: Literacy of four villages: a simple bar diagram.
- Medium: Boys and girls in five divisions: a joint bar diagram.
- Exam type: Comparing the percentage of food, rent, education and other expenses of two families: a percentage bar diagram.
In a percentage bar diagram every bar is always 100% tall. In a sub-divided bar the height depends on the total.
Thinking joint and sub-divided bar diagrams are the same. In a joint bar the parts are side by side; in a sub-divided bar they are stacked.
When finding percent the divisor is always the total, not the other part: 30 ÷ 40, not 30 ÷ 10. Memory clue: Percent: "part ÷ total × 100"; the total always goes below.
Spot the difference
| Sub-divided | Percentage | |
|---|---|---|
| Height of bar | equal to the total | always 100% |
| Parts shown in | actual numbers | percent |
| Used for | seeing the total and parts | comparing shares of parts |
Practice questions
- 1A farm has 25 quintal of wheat and 15 quintal of gram. Scale 1 cm = 10 quintal. How tall is the total bar in cm?
- 2Out of 60 students, 18 walk to school. What percent is that?
- 3Which bar diagram is used to show the percent of the parts?
- 4Year 1: A 12, B 28. Year 2: A 30, B 20. Find the percent of A in each year.
- 5A family earns ₹20,000 and spends ₹8,000 on food. What percent is the food?
- 6In a percentage bar 1 cm = 10%. How many cm is a 65% part? How tall is the full bar?
- 7A shop sold 18 pens and 12 notebooks on Monday, and 30 pens and 20 notebooks on Tuesday. Find the percent of pens on both days and comment on the heights of the sub-divided bars (1 cm = 10).
- 8Total production is 80 quintal and wheat is 45% of it. How many quintal of wheat and jowar? With 1 cm = 10 quintal, how many cm is each part?
Show answers (try them yourself first)
- 125 + 15 = 40 ⇒ 40 ÷ 10 = 4 cm. 2) 18 ÷ 60 × 100 = 30%. 3) The percentage bar diagram.
- 4Year 1: 12 ÷ 40 × 100 = 30%. Year 2: 30 ÷ 50 × 100 = 60%. 5) 8,000 ÷ 20,000 × 100 = 40%. 6) 65% = 6.5 cm. The full bar 100% = 10 cm.
- 7Monday: 18 ÷ 30 × 100 = 60%. Tuesday: 30 ÷ 50 × 100 = 60%. The percents are equal, but the sub-divided bars are 3 cm and 5 cm, so different heights. 8) Wheat 80 × 45 ÷ 100 = 36 quintal. Jowar 80 − 36 = 44 quintal. Heights 3.6 cm and 4.4 cm.
Data, sample and frequency table
If classmates write down daily hours of play, we get scattered numbers in a diary. Making a table with tally marks shows at once how many times each number occurred.
The numbers collected are called data. Each number is a score or observation. The number of times a score occurs is its frequency. The sum of all frequencies, N, is the total number of observations.
In the figure: Above: primary and secondary data. Below: a big group (population) and the small part chosen from it (sample).
Rule 1: Primary and secondary data — Data collected by ourselves is primary. Data collected by others and used by us is secondary.
- Easy: We measured the heights of classmates ourselves: primary data.
- Medium: Census figures taken from a book: secondary data.
- Exam type: (a) We asked neighbours about their electricity bills, (b) rainfall from a weather department report, (c) prices in a newspaper, (d) we counted vehicles passing our gate. Primary: (a), (d). Secondary: (b), (c).
Rule 2: Make a frequency table with tally marks — Write the scores from smallest to largest in one column. Make a tally for each score. The fifth tally is drawn across the earlier four. Finally check the total N.
- Easy: Scores 2, 3, 3, 2, 3: score 2 → ||, f = 2; score 3 → |||, f = 3. N = 2 + 3 = 5.
- Medium: Books read in a month by 20 students: 3, 2, 4, 3, 1, 3, 5, 2, 3, 4, 2, 3, 1, 3, 2, 4, 5, 3, 1, 2. Frequencies: 1 → 3, 2 → 5, 3 → 7, 4 → 3, 5 → 2. Total 3 + 5 + 7 + 3 + 2 = 20.
In the figure: The table of the above 20 numbers with tally marks. The fifth mark is drawn across.
- Exam type: From the table: students who read more than 3 books = 3 + 2 = 5; at most 2 books = 3 + 5 = 8; exactly 3 books = 7.
Rule 3: Population and sample — If collecting data from everyone is hard, we study a randomly chosen sample. If the sample is fair, it represents the whole group.
- Easy: A group of 40 chosen students to know the average height of 800 students: a sample.
- Medium: A handful of rice is checked to see if a sack is good: the handful is the sample and the sack is the population.
- Exam type: A survey of 60 of 2,400 households. The sample is 60 ÷ 2,400 = 1/40 = 2.5% of the households.
N = the sum of all frequencies. Always check N after completing the table.
Counting a number twice or missing it while tallying. If N does not match at the end, count again.
Strike off each number as you tally it so nothing is missed. Memory clue: While reading data use tally marks: four upright and one across.
Spot the difference
| Primary data | Secondary data |
|---|---|
| collected by ourselves | collected by someone else |
| more reliable but takes time | quick but check the source |
| e.g. heights we measured | e.g. census in a book |
Practice questions
- 1We counted the cars passing the school gate. What kind of data is this?
- 2In the scores 4, 5, 4, 4, 5, what is the frequency of 4?
- 3If the frequencies are 5, 8, 7, what is N?
- 4Make a frequency table of the scores 3, 4, 2, 5, 4, 1, 3, 4, 2, 5, 4, 3.
- 5In the above table, how many scored more than 3?
- 6A sample of 50 was taken from 1,000 people. What percent is the sample?
- 7In a table the frequencies of scores 1, 2, 3, 4 are 4, ?, 6, 5 and N = 20. Find the frequency of score 2.
- 8In the same table, how many scored at least 3 and how many at most 2? What percent scored at least 3?
Show answers (try them yourself first)
- 1Primary data. 2) 3. 3) 5 + 8 + 7 = 20.
- 41 → 1, 2 → 2, 3 → 3, 4 → 4, 5 → 2. Total 12. 5) Scores 4 and 5: 4 + 2 = 6. 6) 50 ÷ 1,000 × 100 = 5%.
- 74 + ? + 6 + 5 = 20 ⇒ ? = 20 − 15 = 5. 8) At least 3: 6 + 5 = 11. At most 2: 4 + 5 = 9. Percent 11 ÷ 20 × 100 = 55%.
Grouped frequency distribution
If 50 students have marks anywhere from 0 to 50, a separate row for every mark makes the table very long. So we make groups like 0–10, 10–20, just as school groups children by age.
A group of scores is a class. The ends of a class are the class limits. Upper limit − lower limit = class width. (Lower + upper limit) ÷ 2 = class mark. The number of observations in a class is its frequency.
In the figure: Class 20–30: lower limit 20, upper limit 30, width 10 and class mark 25.
Rule 1: Identify the parts of a class — Width = upper − lower. Class mark = average of the two limits. From the class mark and width we get the class: mark ± half the width.
- Easy: Class 40–45: width 5 and class mark (40 + 45) ÷ 2 = 42.5.
- Medium: Class mark 35 and width 10. Half the width is 5. The class is 35 − 5 = 30 to 35 + 5 = 40, that is 30–40.
- Exam type: The class marks are 12, 17, 22. The difference is 5, so the width is 5. The first class is 12 − 2.5 = 9.5 to 12 + 2.5 = 14.5; the next are 14.5–19.5 and 19.5–24.5.
Rule 2: Exclusive method: the upper limit is not in the class — Classes are written continuously: 0–10, 10–20, 20–30. A score of 10 goes into 10–20, not 0–10. Decimal numbers also fit.
- Easy: In the classes 0–10, 10–20, 20–30, where does the score 10 go? Into 10–20.
- Medium: Marks of 30 students (out of 50), width 10: 0–10 → 4, 10–20 → 5, 20–30 → 6, 30–40 → 6, 40–50 → 9. Total 30.
Grouped frequency distribution
In the app you can move this figure with your finger.
Open in the app- Exam type: The same 30 marks with width 5: 2, 2, 3, 2, 4, 2, 3, 3, 5, 4 (ten classes). Total 30. Changing the width does not change N.
Rule 3: Inclusive method: both limits are in the class — For whole numbers the classes are written like 6–10, 11–15, 16–20. A score of 10 goes into 6–10 and 11 goes into 11–15.
In the figure: See where 15 goes in the two methods. Inclusive classes have a gap between them, exclusive classes do not.
- Easy: Classes 6–10 and 11–15: score 10 → 6–10 and score 11 → 11–15.
- Medium: The class mark of the inclusive class 11–15 is (11 + 15) ÷ 2 = 13.
- Exam type: Scores 7, 12, 15, 16, 20, 9, 13, 18, 11, 6, 10 and classes 6–10, 11–15, 16–20: 6–10 → 6, 7, 9, 10 = 4; 11–15 → 11, 12, 13, 15 = 4; 16–20 → 16, 18, 20 = 3. N = 11.
Keep the class width the same for all classes. 5 or 10 is usually convenient.
In the exclusive method, counting a score equal to the upper limit in the same class. It goes to the next class.
After making the table, check that the frequencies add up to the total number of observations. Memory clue: While counting classes decide first which class a boundary value goes into.
Spot the difference
| Inclusive | Exclusive | |
|---|---|---|
| Classes | 11–15, 16–20 | 10–15, 15–20 |
| Upper limit | is in the class | goes to the next class |
| Between classes | there is a gap | there is no gap |
Practice questions
- 1State the lower and upper limits of the class 35–40.
- 2What is the class mark of 35–40?
- 3What is the width of the class 50–60?
- 4If the class mark is 25 and the width is 10, which is the class?
- 5With the classes 30–40 and 40–50 (exclusive), where does the score 40 go?
- 6Group the scores 5, 12, 18, 21, 9, 15, 29, 22, 10, 20 into the exclusive classes 0–10, 10–20, 20–30.
- 7The class marks are 6, 10, 14. Find the class width and the classes (exclusive).
- 8Heights (cm) of 12 children: 138, 142, 150, 145, 139, 141, 147, 151, 144, 140, 149, 143. Make a frequency table with the inclusive classes 135–139, 140–144, 145–149, 150–154.
Show answers (try them yourself first)
- 1Lower 35 and upper 40. 2) (35 + 40) ÷ 2 = 37.5. 3) 60 − 50 = 10.
- 425 − 5 = 20 to 25 + 5 = 30, that is 20–30. 5) In 40–50 (the upper limit goes to the next class). 6) 0–10: 5, 9 → 2. 10–20: 12, 18, 15, 10 → 4. 20–30: 21, 29, 22, 20 → 4. N = 10.
- 7The difference is 4 ⇒ width 4. Classes: 6 ± 2 = 4–8; 10 ± 2 = 8–12; 14 ± 2 = 12–16. 8) 135–139 → 138, 139 = 2; 140–144 → 142, 141, 144, 140, 143 = 5; 145–149 → 145, 147, 149 = 3; 150–154 → 150, 151 = 2. N = 12.
Cumulative frequency
Children arrive in a line on the school ground and we keep a running total: the first 4, then 9 more make 13, then 15 more make 28. This running total is the cumulative frequency.
The frequency of a class plus all the frequencies of the classes before it is the less-than cumulative frequency (below the upper limit). Subtracting the earlier classes from the total N gives the equal-or-more cumulative frequency (at or above the lower limit).
In the figure: Table of marks of 40 students. Going down, the less-than column is built by adding and the equal-or-more column by subtracting.
Rule 1: Less than: keep adding downwards — The cumulative frequency of the first class equals its own frequency. For the next class = previous cumulative + that class's frequency. The last one must be N.
- Easy: Frequencies 3 and 5: less-than cumulative 3 and 3 + 5 = 8.
- Medium: 30 students: 0–10 → 3, 10–20 → 7, 20–30 → 11, 30–40 → 6, 40–50 → 3. Cumulative: 3, 10, 21, 27, 30. So 21 students scored less than 30.
- Exam type: The 40 students in the figure: 4, 9, 15, 12 ⇒ 4, 13, 28, 40. Students with less than 30 marks: 28.
Rule 2: Equal or more: keep subtracting from N — The first class's cumulative is N. The next = previous − the frequency of the previous class. The last class's cumulative equals its own frequency.
- Easy: N = 8, frequencies 3 and 5: equal-or-more cumulative 8 and 8 − 3 = 5.
- Medium: The same 30 students: 30, 27, 20, 9, 3. So 20 students scored 20 or more and 3 scored 40 or more.
Cumulative frequency
In the app you can move this figure with your finger.
Open in the app- Exam type: 40 students: 40, 40 − 4 = 36, 36 − 9 = 27, 27 − 15 = 12. Students with 20 or more marks: 27.
Rule 3: Read answers from the cumulative table — For "less than x" use the less-than column. For "x or more" use the other column. The number between two limits = the difference of two cumulative values.
- Easy: In the above 40 students, less than 30: 28.
- Medium: 10 or more but less than 30: 28 − 4 = 24 (that is 9 + 15).
- Exam type: The same 24 comes from the other column too: 36 with 10 or more and 12 with 30 or more; 36 − 12 = 24.
The less-than column ends with N and the equal-or-more column starts from N.
Confusing the two columns. Use the upper limit for "less than" and the lower limit for "or more".
Check: for the same number, (less than) + (equal or more) = N. For example, less than 20 is 13 and 20 or more is 27; 13 + 27 = 40. Memory clue: "Less than" goes up, "or more" comes down; both end at N.
Spot the difference
| Less than | Equal or more | |
|---|---|---|
| Start | from the first class f | from N |
| Next value | previous + f | previous − f of the previous class |
| End | N | f of the last class |
Practice questions
- 1Write the less-than cumulative frequencies for the frequencies 5, 8, 7.
- 2N = 20 and the first class has f = 5. What is the equal-or-more cumulative of the second class?
- 3In the 40-student table (4, 9, 15, 12), how many scored less than 20?
- 4Frequencies 6, 10, 14, 5: make both types of cumulative table.
- 5How many observations are in the third class of the above table?
- 6If the equal-or-more cumulative values are 50, 42, 25, 8, find the frequency of each class.
- 7Of 60 students, 18 scored less than 20 and 45 scored less than 40. How many scored between 20 and 40? How many scored 40 or more?
- 8Heights of 30 children: 140–145 → 6, 145–150 → 12, 150–155 → 9, 155–160 → 3. How many are shorter than 155 cm and how many are 150 cm or taller?
Show answers (try them yourself first)
- 15, 13, 20. 2) For the second class: N − f of the first class = 20 − 5 = 15. 3) 4 + 9 = 13.
- 4Less than: 6, 16, 30, 35. Equal or more: 35, 29, 19, 5. 5) 30 − 16 = 14. 6) 8, 17, 17, 8 (N = 50).
- 745 − 18 = 27 children between 20 and 40. 40 or more = 60 − 45 = 15. 8) Less-than cumulative: 6, 18, 27, 30. Shorter than 155: 27. Equal-or-more: 30, 24, 12, 3. 150 or taller: 12.
Mean, median and mode
To say in one number "what are the marks of the class in general?", we use three numbers: the average (mean), the middle number (median) and the most frequent number (mode).
Mean = sum of all observations ÷ number of observations. Median = the number exactly in the middle when arranged in ascending order; if the number of observations is even, the average of the middle two. Mode = the number that occurs most often.
Mean, median and mode
In the app you can move this figure with your finger.
Open in the appRule 1: Mean: sum ÷ count — Add all numbers and divide by how many there are. With a frequency table, add each score × its frequency and divide by N.
In the figure: The mean is the balancing point of a beam. The distances on the two sides of 7 are equal: 9 and 9.
- Easy: 4, 6, 8, 10, 12: sum 40; 40 ÷ 5 = 8.
- Medium: 3, 4, 5, 9, 14: sum 35; 35 ÷ 5 = 7 (as in the figure).
- Exam type: The books table (1 → 3, 2 → 5, 3 → 7, 4 → 3, 5 → 2, N = 20): Σ f × x = 3 + 10 + 21 + 12 + 10 = 56. Mean = 56 ÷ 20 = 2.8 books.
Rule 2: Median: arrange in ascending order first — If the count is odd, take the exact middle. If it is even, take the average of the middle two.
In the figure: Among 7 numbers the 4th is the middle. Among 6 numbers the 3rd and 4th are the middle.
- Easy: 12, 15, 18, 20, 23, 27, 31 (7 numbers): the 4th number is 20 ⇒ median 20.
- Medium: 8, 10, 13, 15, 19, 22 (6 numbers): the middle ones are 13 and 15; (13 + 15) ÷ 2 = 14.
- Exam type: 25, 12, 18, 30, 9, 21: first in ascending order 9, 12, 18, 21, 25, 30. The middle ones are 18 and 21; median (18 + 21) ÷ 2 = 19.5.
Rule 3: Mode: the highest frequency — The number with the highest frequency is the mode. If two numbers share the highest frequency there are two modes. If no number repeats there is no mode.
In the figure: The tallest bar is the mode. Here shoe size 7 occurs the most.
- Easy: Shoe sizes 5, 6, 7, 8, 9 with frequencies 3, 7, 12, 6, 2: the highest is 12, so the mode is 7.
- Medium: 4, 6, 6, 7, 9, 9, 10: 6 and 9 occur twice each; two modes: 6 and 9.
- Exam type: 3, 5, 8, 9: no number repeats; there is no mode.
Numbers must be arranged in ascending order before finding the median. The mean changes because of a very large or small number; the median and mode change less.
Taking the middle number without arranging the numbers, or taking just one number as the median when the count is even.
Check: the mean always lies between the smallest and the largest number. Memory clue: Mean = average, median = middle, mode = most.
Spot the difference
| Mean | Median | Mode | |
|---|---|---|---|
| Meaning | average | middle value in ascending order | most frequent value |
| How to find | sum ÷ count | arrange and take the middle | look at frequencies |
| Effect of a very large value | large | small | none |
Practice questions
- 1What is the mean of 2, 4, 6, 8, 10?
- 2What is the median of 9, 3, 7?
- 3What is the mode of 4, 5, 4, 6, 4?
- 4What is the mean of 12, 15, 18, 20, 10?
- 5What is the median of 14, 8, 20, 10, 6, 16?
- 6Scores 2, 3, 4, 5 have frequencies 4, 6, 6, 4. Find the mean and the mode.
- 7The mean of 6 numbers is 8. Five of them are 5, 7, 9, 10, 6. What is the sixth?
- 8Salaries (₹) of eight workers: 12,000, 12,000, 13,000, 14,000, 15,000, 15,000, 15,000, 60,000. Find the mean, median and mode. Which suits as the usual salary, and why?
Show answers (try them yourself first)
- 1Sum 30; 30 ÷ 5 = 6. 2) In ascending order 3, 7, 9 ⇒ median 7. 3) 4 (three times).
- 4Sum 75; 75 ÷ 5 = 15. 5) In ascending order 6, 8, 10, 14, 16, 20. Median (10 + 14) ÷ 2 = 12. 6) Σ f × x = 8 + 18 + 24 + 20 = 70; N = 20; mean 70 ÷ 20 = 3.5. The frequency 6 occurs twice: modes 3 and 4.
- 7Total 6 × 8 = 48. The five given add to 37. The sixth number is 48 − 37 = 11. 8) Mean 1,56,000 ÷ 8 = 19,500. Median (14,000 + 15,000) ÷ 2 = 14,500. Mode 15,000. The single large salary of 60,000 raises the mean, so the median or mode suits better.